Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        double[] p = new double[n];

        for(int i = 0; i < n; i++){
            p[i] = sc.nextDouble();
        }

        int T = 0;
        int H = 1;
        double[][] a = new double[n][n+1];

        a[0][0] = 1 - p[0];
        a[0][1] = p[0];

        for(int i = 1; i < n; i++){
            for(int j = 0; j <= i+1; j++){
                if(j != 0)
                a[i][j] = p[i]*(a[i-1][j-1]);
                a[i][j] += (1-p[i])*(a[i-1][j]);
            }
        }

        double ans = 0;

        for(int i = (int) Math.ceil(n/2.0); i <= n; i++){
            ans += a[n-1][i];
        }

        System.out.println(ans);
    }
}